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Let $ \sum a_n $ be a positive term series, and let

]]>This content is also available in PDF format.

Let $ \sum a_n $ be a positive term series, and let $ a_n = f(n) $ such that $ f(n) $ decreases as $ n $ increases. Then $ \sum a_n $ converges or diverges if $ \int_1^\infty f(x) dx $ is finite or infinite respectively.

Let $ \sum a_n $ be a positive term series given by $a_n = \frac1{n^p} $. Then, $ \sum a_n $ is convergent if $ p \gt 1 $, and divergent if $ p \leq 1 $.

Let $\sum a_n $ be a positive term series, then:

- $ \sum a_n $ is convergent if $ \sum b_n $ is another convergent series with $ a_n \leq b_n $.
- $ \sum a_n $ is divergent if $ \sum d_n $ is another divergent series with $ a_n \geq d_n $.

Let $ \sum a_n $ and $\sum b_n $ be two positive term series.

- If $ \lim_{n \rightarrow \infty} \frac{a_n}{b_n} $ is a finite and non-zero positive quantity, then $ \sum a_n $ and $\sum b_n$ will converge and diverge together.
- If $ \lim_{n \rightarrow \infty} \frac{a_n}{b_n} = 0 $ and $\sum b_n $ is convergent, then $ \sum a_n $ is also convergent.
- If $\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = \infty $ and $ \sum b_n $ is divergent, then $ \sum a_n $ is also divergent.

Let $ \sum a_n $ be a positive term series, and let $ \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = r $.

- The series is convergent if $ r \lt 1 $.
- The series is divergent if $ r \gt 1 $, or if $r$ is infinite.
- The test fails if $r=1$.

Let $ \sum a_n $ be a positive term series, and $ \lim_{n \rightarrow \infty} (a_n)^{\frac1n} = r $.

1 The series is convergent if $ r \lt 1 $.

2 The series is divergent if $ r \gt 1 $.

3 The test fails if $ r = 1 $.

Let $ \sum a_n $ be a positive term series, and $ \lim_{n \rightarrow \infty} n \left( \frac{a_n}{a_{n+1}}-1 \right) = k$.

- The series is convergent if $ k \gt 1 $.
- The series is divergent if $ k \lt 1 $.
- The test fails if $ k = 1 $.

Let $ \sum a_n $ be a positive term series, and $ \lim_{n \rightarrow \infty} \log\left( \frac{a_n}{a_{n+1}} \right) = k$.

- The series is convergent if $ k \gt 1 $.
- The series is divergent if $ k \lt 1 $.
- The test fails if $ k = 1 $.

If the series $ \sum (-1)^n a_n $ is an alternating series, then the series is convergent if:

- Each term is numerically lesser than the preceeding term. ( $ |a_{n+1}| \lt |a_n| $ )
- $ \lim_{n \rightarrow \infty} a_n = 0 $.

Modular arithmetic is a system of arithmetic in which numbers wrap around, or get ‘reset’ once they reach a certain value. You can think of it as arithmetic using a

]]>Modular arithmetic is a system of arithmetic in which numbers wrap around, or get ‘reset’ once they reach a certain value. You can think of it as arithmetic using a number circle as opposed to a number line.

Consider a circle having circumference $2$ units:

First, notice that if you start at the point marked $0$ and move $2$ units clockwise, you would return to the point marked zero. Also note that, if we start at $0$, moving $3$ units clockwise and $1$ unit clockwise will result in reaching the same destination. We represent this fact using the notation, $ 3 \equiv 1 \pmod 2 $, which is read as “$3$ is congruent to $1$ modulo $2$”. This is equivalent to saying that $3$ and one leave the same remainder when divided by $2$, or that $ (3 - 1) $ is an integral multiple of $2$.

Hence, the statement $ x \equiv y \pmod a $ is equivalent to the following statements:

- $ x-y $ is an integral multiple of $a$, and
- $x$ and $y$ leave the same remainder upon division by $a$.

Modular arithmetic is very useful because of some of the properties of congruence relations. If $ a_1 \equiv b_1 \pmod n $ and $ a_2 \equiv b_2 \pmod n$, then the following properties will hold true.

We will find the last property to be particularly useful.

First, we note that $ 2 \equiv -1 \pmod 3 $.

Because of the last property, this implies that $ 2^{100} \equiv (-1)^{100} \pmod 3 $.

Because $ (-1)^{100} = 1 $, the remainder will be 1. This can be verified using Wolfram|Alpha

First, we note that $ 7^{2010} = 49^{1005} $. Now, because $ 49 \equiv -1 \pmod {25} $,

$$ 49^{1005} \equiv (-1)^{1005} \pmod {25} $$ $$ \Rightarrow 49^{1005} \equiv -1 \pmod {25} $$ $$ \Rightarrow 49^{1005} \equiv 24 \pmod {25} $$

In the last step, I added $25$ to the right side of the congruence. I was able to do so because of the first property.

Hence, the remainder on dividing $ 7^{2010} $ by $25$ is $24$.

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